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i tried this one... any ideas?
My work just disappeared.
I set (a - bi) = to the square of (a + bi).
I had a = a^2 - b^2 and -b = 2ab.
Then, I decided the answers might be 0 + 0i, 1 + 0i, and -1 + 0i but that's just a conjecture.
i didn't check but I got if z = a + bi,
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no, doesn't work.
On the a = a^2, I had that but remembered that all the real parts had to be set equal and the -b^2 is real so a = a^2 - b^2. Then the imaginary parts are set to be equal. so =b = 2ab.
Now, I'm thinking that 0 + 1i and 0 - 1i are solutions.
So far, i'm considering these: 0 + 0i, 1 + 0i, -1 + 0i, and 0 + 1i and 0 - 1i as solutions.
yea. looks good.. 4 solutions
I think there's an elegant way to write a general solution but I don't know what it is.